查询中的 WHERE 子句在 CodeIgniter 3 中无法正常工作

WHERE clause in the query doesn't work properly in CodeIgniter 3

我正在为一个项目使用 CodeIgniter 3,并在控制器中编写了以下函数以将一些数据传递给视图。它查询数据库中的 table 以获取血型作为标签,并获取“isAvailable = 1”处每种血型的行数。然后将这些数据传递到视图中以呈现图表。但如您所见,这些计数是错误的。即使“isAvailable = 0”,它也会对行进行计数。我的代码有什么问题,我该如何解决?

控制器中的函数。

public function bloodTypesChart()
    {

        $chartData = [];
        $blood_types = $this->db->query("SELECT (BloodType) as blood_type FROM packets WHERE (isAvailable) = '1' GROUP BY blood_type")->result_array();

        foreach($blood_types as $bt)
        {
            $record =  $this->db->query("SELECT COUNT(PacketID) as count FROM packets WHERE BloodType = '{$bt['blood_type']}'")->result_array();

            foreach($record as $row) {
                $chartData['label'][] = $bt['blood_type'];
                $chartData['data'][] = $row['count'];
            }
        }
        $chartData['chart_data'] = json_encode($chartData);
        $this->load->view('insight',$chartData);
    }

查看

<script>
    new Chart(document.getElementById("bar-chart"), {
        type: 'bar',
        data: {
            labels: <?= json_encode($label)?>,
            datasets: [
                {
                    label: "Donations",
                    backgroundColor: ["#3e95cd", "#8e5ea2","#3cba9f","#e8c3b9","#c45850"],
                    data: <?= json_encode($data)?>
                }
            ]
        },
        options: {
            legend: { display: false },
            title: {
                display: true,
                text: 'Donations'
            }
        }
    });
</script>

在您的查询中,因为您在括号中封装了 only 字段名称,WHERE (isAvailable) = '1' 的计算结果为 WHERE there is a field labeled isAvailable - 对于table。删除括号,它应该可以正常工作

$blood_types = $this->db->query("SELECT (BloodType) as blood_type FROM packets WHERE isAvailable = '1' GROUP BY blood_type")->result_array();

你这个方法太复杂了,避免使用php循环获取你可以通过简单查询获取的数据。

只需使用此 mysql 查询,计算 isAvailable 行数,当为真 (1) 时:

$sql=" SELECT BloodType as blood_type, COUNT(PacketID) as mycount 
       FROM packets 
       WHERE isAvailable = 1 
       GROUP BY blood_type
     ";

注意:我已将别名 count 更改为 mycount,因为 count 是保留字。

你的函数应该是这样的:

public function bloodTypesChart()
    {
        $sql=" SELECT BloodType as blood_type, COUNT(PacketID) as mycount 
            FROM packets 
            WHERE isAvailable = 1 
            GROUP BY blood_type
         ";    

        $chartData = [];
        $blood_types = $this->db->query($sql)->result_array();

        foreach($blood_types as $row)
        {
                $chartData['label'][] = $row['blood_type'];
                $chartData['data'][] = $row['mycount'];
        }
        $chartData['chart_data'] = json_encode($chartData);
        $this->load->view('insight',$chartData);
    }

这里 sql-fiddle 使用您的数据库的简化版本执行查询